Degenerate Perturbation Theory

Degenerate perturbation theory is the correct first step when a perturbation acts inside a subspace of states with the same unperturbed energy. Ordinary nondegenerate formulas fail because they divide by zero energy differences and because the “right” zeroth-order states are not determined until the perturbation is considered.

The core rule is:

$$\text{Diagonalize }V\text{ inside the degenerate subspace.}$$

Setup

Let $H_0$ have a degenerate eigenspace $\mathcal D$ with energy $E^{(0)}$:

$$H_0\lvert a\rangle = E^{(0)}\lvert a\rangle, \qquad a=1,\ldots,d.$$

Let $P$ be the projector onto this subspace:

$$P=\sum_{a=1}^d\lvert a\rangle\langle a\rvert.$$

The Hamiltonian is

$$H(\lambda)=H_0+\lambda V.$$

Inside $\mathcal D$, define the perturbation matrix

$$W_{ab} = \langle a|V|b\rangle.$$

Equivalently,

$$W=PVP$$

as an operator on the degenerate subspace.

Good Zeroth-Order States

The correct zeroth-order states are eigenvectors of $W$:

$$W\lvert \alpha\rangle = w_\alpha\lvert \alpha\rangle, \qquad \lvert\alpha\rangle\in\mathcal D.$$

Then the first-order energies are

$$E_\alpha = E^{(0)}+\lambda w_\alpha+O(\lambda^2).$$

The perturbation may split the degeneracy if the eigenvalues $w_\alpha$ differ. If some $w_\alpha$ remain degenerate, one must continue the degenerate analysis at the next relevant order or use additional symmetry information.

Why Nondegenerate Formulas Fail

The nondegenerate first-order state correction contains terms

$$\frac{V_{mn}} {E_n^{(0)}-E_m^{(0)}}.$$

For states $m$ inside the degenerate subspace, the denominator is zero. This is not a removable technical problem. It says the perturbation can produce order-$1$ rotations inside the degenerate subspace even when $\lambda$ is small.

The right basis must be chosen before expanding outside the subspace.

Two-State Example

Suppose two states $\lvert1\rangle,\lvert2\rangle$ share energy $E^{(0)}$ and the perturbation matrix is

$$W = \begin{pmatrix} a&b\\ b^*&c \end{pmatrix}.$$

The first-order shifts are the eigenvalues of $W$:

$$w_\pm = \frac{a+c}{2} \pm \sqrt{ \left(\frac{a-c}{2}\right)^2 + \lvert b\rvert^2 }.$$

If $a=c$ and $b\ne0$, the good zeroth-order states are symmetric and antisymmetric combinations up to the phase of $b$, and the degeneracy splits by $2\lvert b\rvert$ at first order.

Role of Symmetry

Symmetry often explains degeneracy and constrains $W$. If $V$ respects the symmetry responsible for the degeneracy, $W$ may be proportional to the identity inside an irreducible multiplet, and the degeneracy may remain at first order.

If $V$ breaks that symmetry, $W$ usually splits the multiplet according to the remaining symmetry.

This is why perturbation theory and symmetry should be used together. Diagonalizing a large perturbation matrix without recognizing block structure is both inefficient and less informative.

Corrections Outside the Degenerate Subspace

After diagonalizing $PVP$, each good zeroth-order state $\lvert\alpha\rangle$ can be corrected by mixing with states outside $\mathcal D$:

$$\lvert\alpha^{(1)}_\perp\rangle = \sum_{r\notin\mathcal D} \frac{\langle r|V|\alpha\rangle} {E^{(0)}-E_r^{(0)}} \lvert r\rangle.$$

This formula is analogous to the nondegenerate state correction, but it is applied only after the degenerate subspace has been diagonalized.

Validity

Degenerate perturbation theory assumes that the degenerate subspace is well separated from states outside it:

$$\left| \frac{\lambda\langle r|V|\alpha\rangle} {E^{(0)}-E_r^{(0)}} \right| \ll1$$

for relevant outside states $r$. If nearby outside states also mix strongly, the subspace should be enlarged.

Common Mistakes

• Applying nondegenerate formulas inside a degenerate subspace.
• Diagonalizing $V$ in the wrong basis while ignoring the projector $P$.
• Forgetting that the perturbation chooses the good zeroth-order linear combinations.
• Assuming degeneracy always splits. Symmetry may protect all or part of it.
• Keeping a subspace too small when nearby levels mix strongly.

Cross-Links

• Time-Independent Perturbation Theory
• Stark Shift in a Two-Level Approximation for the exact finite-gap crossover to dipole-labeled linear branches.
• Nondegenerate Perturbation Theory
• Quasi-Degenerate Perturbation Theory
• Hellmann–Feynman Theorem
• Stark Effect as a Perturbation Example
• Small Parameters and Error Estimates
• Matrix Diagonalization
• Spectral Decomposition
• Symmetry Constraints on Hamiltonians
• Approximate Symmetry
• Accidental Symmetry
• Two-Level System

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. Messiah, Quantum Mechanics, Dover, 1999.

Exercises

1. Diagonalize

$$W = \begin{pmatrix} 0&v\\ v&0 \end{pmatrix}$$

for real $v$ and find the first-order energy shifts.

Solution

The eigenvectors are

$$\frac{1}{\sqrt2} (\lvert1\rangle+\lvert2\rangle)$$

with eigenvalue $v$, and

$$\frac{1}{\sqrt2} (\lvert1\rangle-\lvert2\rangle)$$

with eigenvalue $-v$. Thus the first-order energies are

$$E_\pm = E^{(0)}\pm\lambda v+O(\lambda^2).$$

2. Explain why a perturbation proportional to the identity inside a degenerate subspace does not split the degeneracy at first order.

Solution

If

$$PVP=wI_{\mathcal D},$$

then every vector in the degenerate subspace is an eigenvector of $PVP$ with the same eigenvalue $w$. All states receive the same first-order shift $\lambda w$, so energy differences inside the subspace remain zero at first order.